Delta-star and star-delta conversion

Some three-terminal resistor networks refuse to simplify. A resistor from A to B, another from B to C, another from C to A — a triangle, called delta (Δ) — cannot be combined by plain series or parallel rules, because no two of the three resistors share exactly two terminals with nothing else attached. The same three resistors could instead meet at one common internal point, one leg to each terminal — a star (Y), also called wye. The remarkable fact is that any delta network has an electrically identical star equivalent, and vice versa: swap one for the other and every voltage and current outside the three terminals A, B, C is unchanged. This article gives the conversion formulas, shows why they work, and works two full numerical examples — the standard pre-processing trick that unlocks Kirchhoff's-law, node-voltage and mesh-current analysis on networks that series/parallel reduction alone cannot touch.

The core insight

A delta (triangle) of three resistors between terminals A, B, C can always be replaced by an electrically equivalent star (Y) of three resistors from a common point — and back again — using formulas built from products and sums of the resistances, which turns an otherwise unreducible network into one that plain series/parallel rules can finish.

The core idea

Three resistors connecting three terminals A, B, C can be wired two ways. In the delta (Δ) connection, $R_{AB}$ , $R_{BC}$ , $R_{CA}$ sit directly between each pair of terminals, forming a closed triangle. In the star (Y) connection, $R_A$ , $R_B$ , $R_C$ each run from one terminal to a single common node with no direct path between terminals. Both configurations look identical from outside — same current would flow into each terminal for the same applied terminal voltages — even though the internal wiring, and the internal power dissipated, generally differ. Converting delta to star, each star resistor is the product of the two delta resistors that touch its terminal, divided by the sum of all three delta resistors:

R_A = \frac{R_{AB}R_{CA}}{R_{AB}+R_{BC}+R_{CA}}

where: $R_A$ is the star resistance at terminal A; $R_{AB}$ and $R_{CA}$ are the two delta resistors whose subscripts both contain A; the denominator is the sum of all three delta resistances. $R_B$ and $R_C$ follow the same pattern, cycling the letters (A→B→C→A). Converting star to delta, each delta resistor is the sum of all three pairwise products of the star resistors, divided by the star resistance at the one terminal not in that delta resistor's own subscript:

R_{AB} = \frac{R_AR_B+R_BR_C+R_CR_A}{R_C}

where: $R_{AB}$ is the delta resistance between A and B; the numerator is fixed (the same sum of products for all three delta resistors); the divisor is $R_C$ because C is the terminal excluded from the subscript "AB". $R_{BC}$ and $R_{CA}$ follow by cycling the letters.

Figure

A delta triangle of three resistors between terminals A B and C beside an equivalent star of three resistors meeting at a common centre point N

Derivation AMatching terminal resistances to get the delta-to-star formula

The goal is to derive $R_A = \dfrac{R_{AB}R_{CA}}{R_{AB}+R_{BC}+R_{CA}}$ (and its cyclic partners) from a single requirement: the delta and the star must present exactly the same resistance between every pair of terminals, with the third terminal left open, since that is the only way the two networks can be indistinguishable to anything connected only at A, B, C.

Step 1. Write the resistance between A and B with C open, for both forms. In the star, with C open no current flows through $R_C$ , so the A–B path is simply $R_A + R_B$ in series. In the delta, with C open, current can go directly A→B through $R_{AB}$ , or the long way via C through $R_{CA}$ then $R_{BC}$ , so the two paths are in parallel: $R_{AB} \parallel (R_{CA}+R_{BC})$ .

Step 2. Set the two equal, since the networks must agree at this terminal pair: $R_A+R_B = \dfrac{R_{AB}(R_{CA}+R_{BC})}{R_{AB}+R_{BC}+R_{CA}}$ . Writing $S = R_{AB}+R_{BC}+R_{CA}$ for the sum of all three delta resistors, this is $R_A+R_B = \dfrac{R_{AB}(R_{CA}+R_{BC})}{S}$ .

Step 3. Repeat the same reasoning for the other two terminal pairs (B–C with A open, C–A with B open), giving two more equations of the same shape: $R_B+R_C = \dfrac{R_{BC}(R_{AB}+R_{CA})}{S}$ and $R_C+R_A = \dfrac{R_{CA}(R_{BC}+R_{AB})}{S}$ .

Step 4. Solve the three simultaneous equations for $R_A$ alone. Adding all three equations gives $2(R_A+R_B+R_C) = \dfrac{2(R_{AB}R_{BC}+R_{BC}R_{CA}+R_{CA}R_{AB})}{S}$ , so $R_A+R_B+R_C = \dfrac{R_{AB}R_{BC}+R_{BC}R_{CA}+R_{CA}R_{AB}}{S}$ . Subtracting the Step 3 equation for $R_B+R_C$ from this total isolates $R_A = \dfrac{R_{AB}R_{CA}}{S}$ — the product of the two delta resistors touching A, over the sum of all three. Subtracting the other pairs in turn gives $R_B$ and $R_C$ by the same cyclic pattern, confirming the formula stated above. Verified: substituting $R_{AB}=10$ , $R_{BC}=15$ , $R_{CA}=5$ (so $S=30$ ) gives $R_A=1.67$ , $R_B=5$ , $R_C=2.5$ , matching the worked example below exactly.

Worked example

A delta network has $R_{AB} = 10\,\Omega$ , $R_{BC} = 15\,\Omega$ , $R_{CA} = 5\,\Omega$ . Convert it to its equivalent star, then use that star to find the resistance between A and B with terminal C left open.

Step 1. Sum the delta resistances: $R_{AB}+R_{BC}+R_{CA} = 10+15+5 = 30\,\Omega$ . This sum is the common denominator for all three delta-to-star conversions.

Step 2. Find $R_A$ , the star resistor at the terminal touched by $R_{AB}$ and $R_{CA}$ : $R_A = \dfrac{R_{AB}R_{CA}}{30} = \dfrac{10\times5}{30} = 1.67\,\Omega$ .

Step 3. Find $R_B$ , touched by $R_{AB}$ and $R_{BC}$ : $R_B = \dfrac{R_{AB}R_{BC}}{30} = \dfrac{10\times15}{30} = 5\,\Omega$ . Find $R_C$ , touched by $R_{BC}$ and $R_{CA}$ : $R_C = \dfrac{R_{BC}R_{CA}}{30} = \dfrac{15\times5}{30} = 2.5\,\Omega$ .

Step 4. Now use the star to find the A–B resistance with C open. With C disconnected, no current can flow through $R_C$ , so the only path from A to B is through the two remaining star legs in series: $R_{AB,\text{open C}} = R_A + R_B = 1.67 + 5 = 6.67\,\Omega$ . Answer: the equivalent star has $R_A = 1.67\,\Omega$ , $R_B = 5\,\Omega$ , $R_C = 2.5\,\Omega$ , and the resistance between A and B with C open is $6.67\,\Omega$ .

Competitive stretch

Reverse the direction: a star network has $R_A = 4\,\Omega$ , $R_B = 6\,\Omega$ , $R_C = 3\,\Omega$ . Its equivalent delta, by the same three-step method (sum of pairwise products $= 24+18+12 = 54$ , then divide by the excluded star resistor), works out to $R_{AB} = 54/3 = 18\,\Omega$ , $R_{BC} = 54/4 = 13.5\,\Omega$ , $R_{CA} = 54/6 = 9\,\Omega$ . Checking the A–B resistance with C open two ways confirms equivalence: from the star, $R_A + R_B = 4+6 = 10\,\Omega$ ; from the delta, $R_{AB}$ in parallel with $(R_{CA}+R_{BC})$ gives $\dfrac{18\times22.5}{40.5} = 10\,\Omega$ — the two forms agree exactly, as they must for any terminal pair.

Common traps

Pairing the wrong two delta resistors when computing a star resistor — always use the two delta resistors whose subscripts both contain that terminal's letter (e.g., $R_A$ uses $R_{AB}$ and $R_{CA}$ ).
Dividing by the wrong star resistor in the star-to-delta formula — the divisor is always the star resistor of the terminal excluded from the delta resistor's own subscript (e.g., $R_{AB}$ divides by $R_C$ , not $R_A$ or $R_B$ ).
Assuming internal power dissipation matches between the two forms — only the terminal (external) resistance behaviour is guaranteed equal; the two networks are not identical circuits internally.
Trying to apply series or parallel rules straight to a delta or star sub-network without converting first — that is exactly the situation neither rule can resolve on its own.

Ask the tutor: if you would like help spotting a hidden delta or star sub-network buried inside a larger resistor network, ask the AI tutor to walk through it with you.

Why does the delta-star swap not change anything outside the three terminals?

Because the conversion formulas are built specifically to match the resistance between every pair of terminals, with the third terminal left open, in both forms. There are exactly three such terminal-pair conditions, and solving them together for the unknown resistors is how the formulas arise. Since all three terminal behaviours are matched simultaneously, any circuit connected only at A, B and C sees no difference at all between the delta and the star, no matter what is attached there.

Do I need to memorise separate formulas for R_B, R_C, R_BC and R_CA, or can I derive them from R_A and R_AB?

You do not need to memorise six separate formulas. Both directions follow one repeating pattern: for delta-to-star, take the product of the two delta resistors touching a terminal and divide by the sum of all three; for star-to-delta, take the fixed sum of pairwise products and divide by the star resistor of the excluded terminal. Once you have $R_A$ or $R_{AB}$ correctly, cycling the letters A→B→C→A gives the other two immediately.

When would I actually need this conversion instead of just using Kirchhoff's laws directly?

You can always fall back on Kirchhoff's laws or node-voltage/mesh-current analysis without converting anything, but a delta or star embedded in a larger network often makes those methods more tedious to set up. Converting the awkward triangular or star sub-network first collapses it into resistors that combine by ordinary series/parallel rules, simplifying the rest of the network before you even need Kirchhoff's laws — it is a shortcut, not a replacement, and it is especially useful when the delta/star is just one piece of a bigger circuit.

Delta-star and star-delta conversion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads